Studying Taylor series, I wanted to get a sense for what higher derivatives really express in precise terms using the limit definition of the derivative.

Is this correct?

$$\frac{d^2y}{dx^2} = \lim_{h \to 0} \frac{\big(\lim_{k \to 0} \frac{f(x+h+k)-f(x+h)}{k}\big) - \big(\lim_{g \to 0} \frac{f(x+g)-f(x)}{g}\big)}{h}$$

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    $\begingroup$ yes it's correct, if you meant $\frac{d^2 f}{dx^2} =\frac{d f}{dx}(\frac{d f}{dx})$ and see wiki/Finite_difference#Higher-order_differences $\endgroup$ – reuns Jun 8 '16 at 20:24
  • $\begingroup$ @user1952009 Thanks for the link. I'm having trouble reconciling what is written there and what I have here, notably the $h^2$ they have in the denominator. $\endgroup$ – jeremy radcliff Jun 8 '16 at 20:26
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    $\begingroup$ it is because they let $g = k =h$ and you have to add a $\lim_{h \to 0} $ in front of the whole. now the limit doesn't depend on how $h \to 0$ (and if you let $g\to 0$ first, or $g = h$, or $g=h^2$) only when the function is twice differentiable (that's the definition : the limit exists, whatever how everything $\to 0$) $\endgroup$ – reuns Jun 8 '16 at 20:29
  • $\begingroup$ @user1952009 Ah! Thank you, this is very helpful. Also explains where the $(\Delta x)^2$ and higher degrees of delta $x$ come from in Taylor series. $\endgroup$ – jeremy radcliff Jun 8 '16 at 20:33

You are right, but you could do with an easier expression in one limit: $$ f''(x) = \lim_{h \to 0} \frac{f(x+2h) - 2f(x+h) + f(x)}{h^2} $$

  • $\begingroup$ Thank you, this makes sense now with @user1952009's comment. $\endgroup$ – jeremy radcliff Jun 8 '16 at 20:38
  • $\begingroup$ I fear that a coefficient $2$ is missing for the central term $f(x+h)$ as may be deduced from $\dfrac {\frac{f(x+2h)-f(x+h)}h-\frac{f(x+h)-f(x)}h}h$. The second order central difference (as linked by user1952009) $\dfrac {\frac{f(x+h)-f(x)}h-\frac{f(x)-f(x-h)}h}h=\dfrac{f(x+h)-2f(x)+f(x-h)}{h^2}\,$ may be better in practice (as well as $\dfrac{f\left(x+\frac h2\right)-f\left(x-\frac h2\right)}h$ for the derivative). $\endgroup$ – Raymond Manzoni Jun 8 '16 at 21:31
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    $\begingroup$ @RaymondManzoni indeed, a typo, corrected. thank you. You are right, centralized differences perform better, but here i chose this since it flows out of OP's formulation with $h=k$ $\endgroup$ – gt6989b Jun 8 '16 at 23:29
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    $\begingroup$ It should be pointed out that this is true only if the second derivative exists. The limit you give may exist even though the function is not twice differentiable at $x$. $\endgroup$ – symplectomorphic Jun 9 '16 at 5:35

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